Two parallel shafts whose centre lines are 4.8 m apart are connected by open belt drive. The diameter of larger pulley is 1.5 m and that of smaller pulley 1 m. The initial tension in the belt when stationary is 3 kN. The mass of the belt is 1.5 kg/m lengt

Two parallel shafts whose centre lines are 4.8 m apart are connected by open belt
drive. The diameter of larger pulley is 1.5 m and that of smaller pulley 1 m. the initial
tension in the belt when stationary is 3 KN. The mass of the belt is 1.5 Kg/m length.
The coefficient of friction between belt and pulley is 0.3. Taking centrifugal tension
in to account, calculate power transmitted when smaller pulley rotates at 400 rpm.

Given Data:
Open Belt Drive:
Where,
C = 4.8 m
D 1 = 1.5 m
D 2 = 1 m
N 2 = 400 rpm
3
Ti = 3 KN = 3 x 10 N
m = 1.5 Kg/m length
μ = 0.3
Considering Centrifugal Tension (Tc) =mV 2
[1] We know that, Velocity (V) of the Open Belt Drive;

[2] Centrifugal Tension in Belt (Tc),
(T c ) =mV 2 = 1.5 x (21) 2 = 661.5 N

[3] We know that initial tension in Belt (T i ) as,
Let, T 1 = Tension in Tight Side (N)
T 2 = Tension in Slack Side (N)

[3] We know that initial tension in Belt (T i ) as,
Let, T 1 = Tension in Tight Side (N)
T 2 = Tension in Slack Side (N)
T i =
3000 x 2 = T 1 + T 2 + 2 x (661.5)
T 1 + T 2 = 4677 N ....................................Eq. [1]

[4] For an Open Belt Drive,

So, angle of lap on the smaller pulley is;